# RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.1 Solutions

On this page you will find Maths RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.1 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 5 for class 9 deals with the topic of triangles and it is one of the most important chapters.

## Download RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.1 Solutions in PDF

Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 5 Factorisation of Algebraic Expressions Exercise 5.1 below and it will be beneficial for them.

## RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.1 Solutions

Q1. $x^{3}+x-3 x^{2}-3$
SOLUTION:
Taking $x$ common in $x^{3}+x$
$=x\left(x^{2}+1\right)-3 x^{2}-3$
Taking $-3$ common in $-3 \mathrm{x}^{2}-3$
$=\mathrm{x}\left(\mathrm{x}^{2}+1\right)-3\left(\mathrm{x}^{2}+1\right)$
Now, we take $\left(\mathrm{x}^{2}+1\right)$ common
$=\left(x^{2}+1\right)(x-3)$
$\therefore \mathrm{x}^{3}+\mathrm{x}-3 \mathrm{y}^{2}-3=\left(\mathrm{x}^{2}+1\right)(\mathrm{x}-3)$

$Q 2 \cdot a(a+b)^{3}-3 a^{2} b(a+b)$
SOLUTION :
Taking $(a+b)$ common in the two terms
$=(a+b)\left\{a(a+b)^{2}-3 a^{2} b\right\}$
Now, using $(a+b)^{2}=a^{2}+b^{2}+2 a b$
$=(a+b)\left\{a\left(a^{2}+b^{2}+2 a b\right)-3 a^{2} b\right\}$
$=(a+b)\left\{a^{3}+a b^{2}+2 a^{2} b-3 a^{2} b\right\}$
$=(a+b)\left\{a^{3}+a b^{2}-a^{2} b\right\}$
$=(a+b) p\left\{a^{2}+b^{2}-a b\right\}$
$=p(a+b)\left(a^{2}+b^{2}-a b\right)$
$\therefore a(a+b)^{3}-3 a^{2} b(a+b)=a(a+b)\left(a^{2}+b^{2}-a b\right)$

Q3. $x\left(x^{3}-y^{3}\right)+3 x y(x-y)$
SOLUTION:
Elaborating $x^{3}-y^{3}$ using the identity $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$
$=x(x-y)\left(x^{2}+x y+y^{2}\right)+3 x y(x-y)$
Taking common $x(x-y)$ in both the terms
$=x(x-y)\left(x^{2}+x y+y^{2}+3 y\right)$
$\therefore x\left(x^{3}-y^{3}\right)+3 x y(x-y)=x(x-y)\left(x^{2}+x y+y^{2}+3 y\right)$

Q4. $a^{2} x^{2}+\left(a x^{2}+1\right) x+a$
SOLUTION:
We multiply $x\left(a x^{2}+1\right)=a x^{3}+x$
$=a^{2} x^{2}+a x^{3}+x+a$
Taking common $\mathrm{ax}^{2}$ in $\left(\mathrm{a}^{2} \mathrm{x}^{2}+\mathrm{ax}^{3}\right)$ and 1 in $(\mathrm{x}+\mathrm{a})$
$=\mathrm{ax}^{2}(\mathrm{a}+\mathrm{x})+1(\mathrm{x}+\mathrm{a})$
$=\mathrm{ax}^{2}(\mathrm{a}+\mathrm{x})+1(\mathrm{a}+\mathrm{x})$
Taking $(a+x)$ common in both the terms
$=(a+x)\left(a x^{2}+1\right)$
$\therefore \mathrm{a}^{2} \mathrm{x}^{2}+\left(\mathrm{ax}^{2}+1\right) \mathrm{x}+\mathrm{a}=(\mathrm{a}+\mathrm{x})\left(\mathrm{ax}^{2}+1\right)$

Q5. $x^{2}+y-x y-x$
SOLUTION :
On rearranging
$x^{2}-x y-x+y$
Taking $x$ common in the $\left(x^{2}-x y\right)$ and $-1$ in $(-x+y)$
$=x(x-y)-1(x-y)$
Taking $(x-y)$ common in the terms
$=(x-y)(x-1)$
latex] $:[$ /latex $] x^{2}+y-x y-x=(x-y)(x-1)$

Q6 $x^{3}-2 x^{2} b+3 x y^{2}-6 y^{3}$
SOLUTION:
Taking $\mathrm{x}^{2}$ common in $\left(\mathrm{x}^{3}-2 \mathrm{x}^{2} \mathrm{y}\right)$ and $+3 \mathrm{y}^{2}$ common in $\left(3 \mathrm{xy}^{2}-6 \mathrm{y}^{3}\right)$
$=x^{2}(x-2 y)+3 y^{2}(x-2 y)$
Taking $(x-2 y)$ common in the terms
$=(x-2 y)\left(x^{2}+3 y^{2}\right)$
latex]: $[/$ latex $] x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3}=(x-2 y)\left(x^{2}+3 y^{2}\right)$

Q7. $6 a b-b^{2}+12 a c-2 b c$
SOLUTION :
Taking b common in $\left(6 a b-b^{2}\right)$ and $2 c$ in ( $\left.12 a c-2 b c\right)$
$=b(6 a-b)+2 c(6 a-b)$
Taking $(6 a-b)$ common in the terms
$=(6 a-b)(b+2 c)$
latex] $\therefore[/$ latex $] 6 \mathrm{ab}-\mathrm{b}^{2}+12 \mathrm{ac}-2 \mathrm{bc}=(6 \mathrm{a}-\mathrm{b})(\mathrm{b}+2 \mathrm{c})$

Q8. $\left[\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right]-4\left[\mathrm{x}+\frac{1}{\mathrm{x}}\right]+6$
SOLUTION:
$=x^{2}+\frac{1}{x^{2}}-4 x-\frac{4}{x}+4+2$
$=x^{2}+\frac{1}{x^{2}}+4+2-\frac{4}{x}-4 x$
$=\left(x^{2}\right)+\left(\frac{1}{x}\right)^{2}+(-2)^{2}+2 \times x \times \frac{1}{x}+2 \times \frac{1}{x} \times(-2)+2(-2) x$
Using identity
$x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x=(x+y+z)^{2}$
We get,
$=\left[x+\frac{1}{x}+(-2)\right]^{2}$
$=\left[x+\frac{1}{x}-2\right]^{2}$
$=\left[\mathrm{x}+\frac{1}{\mathrm{x}}-2\right]\left[\mathrm{x}+\frac{1}{\mathrm{x}}-2\right]$
$\therefore\left[\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right]-4\left[\mathrm{x}+\frac{1}{\mathrm{x}}\right]+6=\left[\mathrm{x}+\frac{1}{\mathrm{x}}-2\right]\left[\mathrm{x}+\frac{1}{\mathrm{x}}-2\right]$

Q9. $x(x-2)(x-4)+4 x-8$
SOLUTION :
$=x(x-2)(x-4)+4(x-2)$
Taking $(x-2)$ common in both the terms
$=(x-2)\{x(x-4)+4\}$
$=(x-2)\left\{x^{2}-4 x+4\right\}$
Now splitting the middle term of $\mathrm{x}^{2}-4 \mathrm{x}+4$
$=(x-2)\left\{x^{2}-2 x-2 x+4\right\}$
$=(x-2)\{x(x-2)-2(x-2)\}$
$=(x-2)\{(x-2)(x-2)\}$
$=(x-2)(x-2)(x-2)$
$=(x-2)^{3}$
$\therefore x(x-2)(x-4)+4 x-8=(x-2)^{3}$

Q10 $\cdot(x+2)\left(x^{2}+25\right)-10 x^{2}-20 x$
SOLUTION:
$(x+2)\left(x^{2}+25\right)-10 x(x+2)$
Taking $(x+2)$ common in both the terms
$=(x+2)\left(x^{2}+25-10 x\right)$
$=(x+2)\left(x^{2}-10 x+25\right)$
Splitting the middle term of $\left(\mathrm{x}^{2}-10 \mathrm{x}+25\right)$
$=(x+2)\left(x^{2}-5 x-5 x+25\right)$
$=(x+2)\{x(x-5)-5(x-5)\}$
$=(x+2)(x-5)(x-5)$
$\therefore(x+2)\left(x^{2}+25\right)-10 x^{2}-20 x=(x+2)(x-5)(x-5)$

Q11 $.2 \mathrm{a}^{2}+2 \sqrt{6} \mathrm{ab}+3 \mathrm{~b}^{2}$
SOLUTION :
$=(\sqrt{2} a)^{2}+2 \times \sqrt{2} a \times \sqrt{3} b+(\sqrt{3} b)^{2}$
Using the identity $(\mathrm{p}+\mathrm{q})^{2}=\mathrm{p}^{2}+\mathrm{q}^{2}+2 \mathrm{pq}$
$=(\sqrt{2} \mathrm{a}+\sqrt{3} \mathrm{~b})^{2}$
$=(\sqrt{2} a+\sqrt{3} b)(\sqrt{2} a+\sqrt{3} b)$
$\therefore 2 \mathrm{a}^{2}+2 \sqrt{6} \mathrm{ab}+3 \mathrm{~b}^{2}=(\sqrt{2} \mathrm{a}+\sqrt{3} \mathrm{~b})(\sqrt{2} \mathrm{a}+\sqrt{3} \mathrm{~b})$

Q12 $\cdot(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c) \times(b-c+a)$
SOLUTION:
Let $(a-b+c)=x$ and $(b-c+a)=y$
$=x^{2}+y^{2}+2 x y$
Using the identity $(\mathrm{a}+\mathrm{b})^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab}$
$=(\mathrm{x}+\mathrm{y})^{2}$
Now , substituting $x$ and $y$
$(a-b+c+b-c+a)^{2}$
Cancelling $-b,+b \&+c,-c$
$=(2 \mathrm{a})^{2}$
$=4 \mathrm{a}^{2}$
$\therefore(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c) \times(b-c+a)=4 a^{2}$

Q13 $\cdot a^{2}+b^{2}+2(a b+b c+c a)$
SOLUTION :
$=a^{2}+b^{2}+2 a b+2 b c+2 c a$
Using the identity $(\mathrm{p}+\mathrm{q})^{2}=\mathrm{p}^{2}+\mathrm{q}^{2}+2 \mathrm{pq}$
We get,
$=(a+b)^{2}+2 b c+2 c a$
$=(a+b)^{2}+2 c(b+a)$
or $(a+b)^{2}+2 c(a+b)$
Taking ( $a+b$ ) common
$=(a+b)(a+b+2 c)$
$\therefore a^{2}+b^{2}+2(a b+b c+c a)=(a+b)(a+b+2 c)$

Q14. $4(\mathrm{x}-\mathrm{y})^{2}-12(\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y})+9(\mathrm{x}+\mathrm{y})^{2}$
SOLUTION :
Let $(x-y)=x,(x+y)=y$
$=4 x^{2}-12 x y+9 y^{2}$
Splitting the middle term $-12=-6-6$ also $4 \times 9=-6 \times-6$
$=4 x^{2}-6 x y-6 x y+9 y^{2}$
$=2 x(2 x-3 y)-3 y(2 x-3 y)$
$=(2 x-3 y)(2 x-3 y)$
$=(2 x-3 y)^{2}$
Substituting $x=x-y \& y=x+y$
$=[2(x-y)-3(x+y)]^{2}=[2 x-2 y-3 x-3 y]^{2}$
$=(2 x-3 x-2 y-3 y)^{2}$
$=[-x-5 y]^{2}$
$=[(-1)(x+5 y)]^{2}$
$=(x+5 y)^{2}$ $\left[\because(-1)^{2}=1\right]$
$\therefore 4(x-y)^{2}-12(x-y)(x+y)+9(x+y)^{2}=(x+5 y)^{2}$

Q15. $a^{2}-b^{2}+2 b c-c^{2}$
SOLUTION :
$a^{2}-\left(b^{2}-2 b c+c^{2}\right)$
Using the identity $(\mathrm{a}-\mathrm{b})^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab}$
$=a^{2}-(b-c)^{2}$
Using the identity $a^{2}-b^{2}=(a+b)(a-b)$
$=(a+b-c)(a-(b-c))$
$=(a+b-c)(a-b+c)$
$\therefore a^{2}-b^{2}+2 b c-c^{2}=(a+b-c)(a-b+c)$

Q16. $a^{2}+2 a b+b^{2}-c^{2}$
SOLUTION :
Using the identity $(\mathrm{p}+\mathrm{q})^{2}=\mathrm{p}^{2}+\mathrm{q}^{2}+2 \mathrm{pq}$
$=(a+b)^{2}-c^{2}$
Using the identity $\mathrm{p}^{2}-\mathrm{q}^{2}=(\mathrm{p}+\mathrm{q})(\mathrm{p}-\mathrm{q})$
$=(a+b+c)(a+b-c)$
$\therefore \mathrm{a}^{2}+2 \mathrm{ab}+\mathrm{b}^{2}-\mathrm{c}^{2}=(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{a}+\mathrm{b}-\mathrm{c})$

Q17.$a^{2}+4 b^{2}-4 a b-4 c^{2}$
SOLUTION:
On rearranging
$=a^{2}-4 a b+4 b^{2}-4 c^{2}$
$=(a)^{2}-2 \times a \times 2 b+(2 b)^{2}-4 c^{2}$
Using the identity $(a-b)^{2}=a^{2}+b^{2}-2 a b$
$=(a-2 b)^{2}-4 c^{2}$
$=(a-2 b)^{2}-(2 c)^{2}$
Using the identity $a^{2}-b^{2}=(a+b)(a-b)$
$=(a-2 b-2 c)(a-2 b+2 c)$
$\therefore a^{2}+4 b^{2}-4 a b-4 c^{2}=(a-2 b-2 c)(a-2 b+2 c)$

Q18. $x y^{9}-y x^{9}$
SOLUTION:
$=x y\left(y^{8}-x^{8}\right)$
$=x y\left(\left(y^{4}\right)^{2}-\left(x^{4}\right)^{2}\right)$
Using the identity $\mathrm{p}^{2}-\mathrm{q}^{2}=(\mathrm{p}+\mathrm{q})(\mathrm{p}-\mathrm{q})$
$=x y\left(y^{4}+x^{4}\right)\left(y^{4}-x^{4}\right)$
$=x y\left(y^{4}+x^{4}\right)\left(\left(y^{2}\right)^{2}-\left(x^{2}\right)^{2}\right)$
Using the identity $\mathrm{p}^{2}-\mathrm{q}^{2}=(\mathrm{p}+\mathrm{q})(\mathrm{p}-\mathrm{q})$
$=x y\left(y^{4}+x^{4}\right)\left(y^{2}+x^{2}\right)\left(y^{2}-x^{2}\right)$
$=x y\left(y^{4}+x^{4}\right)\left(y^{2}+x^{2}\right)(y+x)(y-x)$
$=x y\left(x^{4}+y^{4}\right)\left(x^{2}+y^{2}\right)(x+y)(-1)(x-y)$
$\because(y-x)=-1(x-y)$
$=-x y\left(x^{4}+y^{4}\right)\left(x^{2}+y^{2}\right)(x+y)(x-y)$
$\therefore x y^{9}-y x^{9}=-x y\left(x^{4}+y^{4}\right)\left(x^{2}+y^{2}\right)(x+y)(x-y)$

Q19. $x^{4}+x^{2} y^{2}+y^{4}$
SOLUTION:
Adding $x^{2} y^{2}$ and subtracting $x^{2} y^{2}$ to the given equation
$=x^{4}+x^{2} y^{2}+y^{4}+x^{2} y^{2}-x^{2} y^{2}$
$=x^{4}+2 x^{2} y^{2}+y^{4}-x^{2} y^{2}$
$=\left(x^{2}\right)^{2}+2 \times x^{2} \times y^{2}+\left(y^{2}\right)^{2}-(x y)^{2}$
Using the identity $(\mathrm{p}+\mathrm{q})^{2}=\mathrm{p}^{2}+\mathrm{q}^{2}+2 \mathrm{pq}$
$=\left(x^{2}+y^{2}\right)^{2}-(x y)^{2}$
Using the identity $\mathrm{p}^{2}-\mathrm{q}^{2}=(\mathrm{p}+\mathrm{q})(\mathrm{p}-\mathrm{q})$
$=\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)$
$\therefore x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)$

Q 20 $x^{2}-y^{2}-4 x z+4 z^{2}$
SOLUTION:
On rearranging the terms
$=x^{2}-4 x z+4 z^{2}-y^{2}$
$=(x)^{2}-2 \times x \times 2 z+(2 z)^{2}-y^{2}$
Using the identity $x^{2}-2 x y+y^{2}=(x-y)^{2}$
$=(x-2 z)^{2}-y^{2}$
Using the identity $\mathrm{p}^{2}-\mathrm{q}^{2}=(\mathrm{p}+\mathrm{q})(\mathrm{p}-\mathrm{q})$
$=(\mathrm{x}-2 \mathrm{z}+\mathrm{y})(\mathrm{x}-2 \mathrm{z}-\mathrm{y})$
$\therefore \mathrm{x}^{2}-\mathrm{y}^{2}-4 \mathrm{xz}+4 \mathrm{z}^{2}=(\mathrm{x}-2 \mathrm{z}+\mathrm{y})(\mathrm{x}-2 \mathrm{z}-\mathrm{y})$

Q21. $x^{2}+6 \sqrt{2} x+10$
SOLUTION :
Splitting the middle term,
$=x^{2}+5 \sqrt{2} x+\sqrt{2} x+10$ $[\because 6 \sqrt{2}=5 \sqrt{2}+\sqrt{2}$ and $5 \sqrt{2} \times \sqrt{2}=10]$
$=\mathrm{x}(\mathrm{x}+5 \sqrt{2})+\sqrt{2}(\mathrm{x}+5 \sqrt{2})$
$=(\mathrm{x}+5 \sqrt{2})(\mathrm{x}+\sqrt{2})$
$\therefore \mathrm{x}^{2}+6 \sqrt{2} \mathrm{x}+10=(\mathrm{x}+5 \sqrt{2})(\mathrm{x}+\sqrt{2})$

Q22. $x^{2}-2 \sqrt{2} x-30$
SOLUTION :
Splitting the middle term,
$=x^{2}-5 \sqrt{2} x+3 \sqrt{2} x-30$
$[\because-2 \sqrt{2}=-5 \sqrt{2}+3 \sqrt{2}$ also $-5 \sqrt{2} \times 3 \sqrt{2}=-30]$
$=x(x-5 \sqrt{2})+3 \sqrt{2}(x-5 \sqrt{2})$
$=(x-5 \sqrt{2})(x+3 \sqrt{2})$
$\therefore \mathrm{x}^{2}-2 \sqrt{2} \mathrm{x}-30=(\mathrm{x}-5 \sqrt{2})(\mathrm{x}+3 \sqrt{2})$

Q23. $x^{2}-\sqrt{3} x-6$
SOLUTION :
Splitting the middle term,
$=x^{2}-2 \sqrt{3} x+\sqrt{3} x-6$ $[\because-\sqrt{3}=-2 \sqrt{3}+\sqrt{3}$ also $-2 \sqrt{3} \times \sqrt{3}=-6]$
$=x(x-2 \sqrt{3})+\sqrt{3}(x-2 \sqrt{3})$
$=(\mathrm{x}-2 \sqrt{3})(\mathrm{x}+\sqrt{3})$
$\therefore x^{2}-\sqrt{3} x-6=(x-2 \sqrt{3})(x+\sqrt{3})$

Q24. $x^{2}+5 \sqrt{5} x+30$
SOLUTION:
Splitting the middle term,
$=x^{2}+2 \sqrt{5} x+3 \sqrt{5} x+30$ $[\because 5 \sqrt{5}=2 \sqrt{5}+3 \sqrt{5}$ also $2 \sqrt{5} \times 3 \sqrt{5}=30]$
$=x(x+2 \sqrt{5})+3 \sqrt{5}(x+2 \sqrt{5})$
$=(x+2 \sqrt{5})(x+3 \sqrt{5})$
$\therefore \mathrm{x}^{2}+5 \sqrt{5} \mathrm{x}+30=(\mathrm{x}+2 \sqrt{5})(\mathrm{x}+3 \sqrt{5})$

Q25. $x^{2}+2 \sqrt{3} x-24$
SOLUTION:
Splitting the middle term,
$=x^{2}+4 \sqrt{3} x-2 \sqrt{3} x-24$ $[\because 2 \sqrt{3}=4 \sqrt{3}-2 \sqrt{3}$ also $4 \sqrt{3}(-2 \sqrt{3})=-24]$
$=x(x+4 \sqrt{3})-2 \sqrt{3}(x+4 \sqrt{3})$
$=(x+4 \sqrt{3})(x-2 \sqrt{3})$
$\therefore \mathrm{x}^{2}+2 \sqrt{3} \mathrm{x}-24=(\mathrm{x}+4 \sqrt{3})(\mathrm{x}-2 \sqrt{3})$

Q26. $2 \mathrm{x}^{2}-\frac{5}{6} \mathrm{x}+\frac{1}{12}$
SOLUTION:
Splitting the middle term,
$=2 x^{2}-\frac{x}{2}-\frac{x}{3}+\frac{1}{12}$ $\left[\because-\frac{5}{6}=-\frac{1}{2}-\frac{1}{3}\right.$ also $\left.-\frac{1}{2} \times-\frac{1}{3}=2 \times \frac{1}{12}\right]$
$=\mathrm{x}\left(2 \mathrm{x}-\frac{1}{2}\right)-\frac{1}{6}\left(2 \mathrm{x}-\frac{1}{2}\right)$
$=\left(2 \mathrm{x}-\frac{1}{2}\right)\left(\mathrm{x}-\frac{1}{6}\right)$
$\therefore 2 \mathrm{x}^{2}-\frac{5}{6} \mathrm{x}+\frac{1}{12}=\left(2 \mathrm{x}-\frac{1}{2}\right)\left(\mathrm{x}-\frac{1}{6}\right)$

Q 27 $x^{2}+\frac{12}{35} x+\frac{1}{35}$
SOLUTION :
Splitting the middle term,
$=x^{2}+\frac{5}{35} x+\frac{7}{35} x+\frac{1}{35}$ $\left[\because \frac{12}{35}=\frac{5}{35}+\frac{7}{35}\right.$ and $\left.\frac{5}{35} \times \frac{7}{35}=\frac{1}{35}\right]$
$=x^{2}+\frac{x}{7}+\frac{x}{5}+\frac{1}{35}$
$=x\left(x+\frac{1}{7}\right)+\frac{1}{5}\left(x+\frac{1}{7}\right)$
$=\left(\mathrm{x}+\frac{1}{7}\right)\left(\mathrm{x}+\frac{1}{5}\right)$
$\therefore \mathrm{x}^{2}+\frac{12}{35} \mathrm{x}+\frac{1}{35}=\left(\mathrm{x}+\frac{1}{7}\right)\left(\mathrm{x}+\frac{1}{5}\right)$

Q28. $21 \mathrm{x}^{2}-2 \mathrm{x}+\frac{1}{21}$
SOLUTION:
$=(\sqrt{21 x})^{2}-2 \sqrt{21} x \times \frac{1}{\sqrt{21}}+\left(\frac{1}{\sqrt{21}}\right)^{2}$
Using the identity $(\mathrm{x}-\mathrm{y})^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{xy}$
$=\left(\sqrt{21} \mathrm{x}-\frac{1}{\sqrt{21}}\right)^{2}$
$\therefore 21 \mathrm{x}^{2}-2 \mathrm{x}+\frac{1}{21}=\left(\sqrt{21} \mathrm{x}-\frac{1}{\sqrt{21}}\right)^{2}$

Q 29 $5 \sqrt{5} x^{2}+20 x+3 \sqrt{5}$
SOLUTION:
Splitting the middle term,
$=5 \sqrt{5} x^{2}+15 x+5 x+3 \sqrt{5}$ $[\because 20=15+5$ and $15 \times 5=5 \sqrt{5} \times 3 \sqrt{5}]$
$=5 x(\sqrt{5} x+3)+\sqrt{5}(\sqrt{5} x+3)$
$=(\sqrt{5} x+3)(5 x+\sqrt{5})$
$\therefore 5 \sqrt{5} x^{2}+20 x+3 \sqrt{5}=(\sqrt{5} x+3)(5 x+\sqrt{5})$

Q30. $2 x^{2}+3 \sqrt{5} x+5$
SOLUTION:
Splitting the middle term,
$=2 x^{2}+2 \sqrt{5} x+\sqrt{5} x+5$
$=2 x(x+\sqrt{5})+\sqrt{5}(x+\sqrt{5})$
$=(x+\sqrt{5})(2 x+\sqrt{5})$
$\therefore 2 x^{2}+3 \sqrt{5} x+5=(x+\sqrt{5})(2 x+\sqrt{5})$

Q31 $.9(2 \mathrm{a}-\mathrm{b})^{2}-4(2 \mathrm{a}-\mathrm{b})-13$
SOLUTION:
Let $2 a-b=x$
$=9 x^{2}-4 x-13$
Splitting the middle term,
$=9 x^{2}-13 x+9 x-13$
$=x(9 x-13)+1(9 x-13)$
$=(9 x-13)(x+1)$
Substituting $x=2 a-b$
$=[9(2 a-b)-13](2 a-b+1)$
$=(18 a-9 b-13)(2 a-b+1)$
$\therefore 9(2 \mathrm{a}-\mathrm{b})^{2}-4(2 \mathrm{a}-\mathrm{b})-13=(18 \mathrm{a}-9 \mathrm{~b}-13)(2 \mathrm{a}-\mathrm{b}+1)$

Q 32 . $7(x-2 y)^{2}-25(x-2 y)+12$
SOLUTION:
Let $x-2 y=P$
$=7 P^{2}-25 P+12$
Splitting the middle term,
$=7 \mathrm{P}^{2}-21 \mathrm{P}-4 \mathrm{P}+12$
$=7 \mathrm{P}(\mathrm{P}-3)-4(\mathrm{P}-3)$
$=(\mathrm{P}-3)(7 \mathrm{P}-4)$
Substituting $\mathrm{P}=\mathrm{x}-2 \mathrm{y}$
$=(x-2 y-3)(7(x-2 y)-4)$
$=(x-2 y-3)(7 x-14 y-4)$
$\therefore 7(x-2 y)^{2}-25(x-2 y)+12=(x-2 y-3)(7 x-14 y-4)$

Q33. $2(x+y)^{2}-9(x+y)-5$
SOLUTION:
Let $x+y=z$
$=2 z^{2}-9 z-5$
Splitting the middle term,
$=2 z^{2}-10 z+z-5$
$=2 z(z-5)+1(z-5)$
$=(z-5)(2 z+1)$
Substituting $z=x+y$
$=(x+y-5)(2(x+y)+1)$
$=(x+y-5)(2 x+2 y+1)$
$\therefore 2(x+y)^{2}-9(x+y)-5=(x+y-5)(2 x+2 y+1)$

Q34. Give the possible expression for the length \& breadth of the rectangle having $35 \mathrm{y}^{2}-13 \mathrm{y}-12$ as its area.
SOLUTION :
Area is given as $35 \mathrm{y}^{2}-13 \mathrm{y}-12$
Splitting the middle term,
Area $=35 \mathrm{y}^{2}+218 \mathrm{y}-15 \mathrm{y}-12$
$=7 \mathrm{y}(5 \mathrm{y}+4)-3(5 \mathrm{y}+4)$
$=(5 \mathrm{y}+4)(7 \mathrm{y}-3)$
We also know that area of rectangle $=$ length $\times$ breadth
$\therefore$ Possible length $=(5 \mathrm{y}+4)$ and breadth $=(7 \mathrm{y}-3)$
Or possible length $=(7 y-3)$ and breadth= $(5 y+4)$

Q35. What are the possible expression for the cuboid having volume $3 \mathrm{x}^{2}-12 \mathrm{x}$.
SOLUTION:
Volume $=3 x^{2}-12 x$
$=3 x(x-4)$
$=3 \times x(x-4)$
Also volume $=$ Length $\times$ Breadth $\times$ Height
$\therefore$ Possible expression for dimensions of cuboid are $=3, x,(x-4)$

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